Does cats mapN run all the futures in parallel? - scala

Question

As mentioned in the jump-start guide, mapN will run all the futures in parallel, so I created the below simple Scala program, but a sample run shows diff to be 9187 ms and diffN to be 9106 ms. So it looks like that the mapN is also running the futures sequentially, isn't it? Please let me know if I am missing something?
package example
import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global
import java.time.LocalDateTime
import java.time.Duration
import scala.util.Failure
import scala.util.Success
import java.time.ZoneOffset
import cats.instances.future._
import cats.syntax.apply._
object FutureEx extends App {
val before = LocalDateTime.now()
val sum = for {
a <- getA
b <- getB
c <- getC
} yield (a + b + c)
sum onComplete {
case Failure(ex) => println(s"Error: ${ex.getMessage()}")
case Success(value) =>
val after = LocalDateTime.now()
println(s"before: $before")
println(s"after: $after")
val diff = getDiff(before, after)
println(s"diff: $diff")
println(s"sum: $value")
}
// let the above finish
Thread.sleep(20000)
val beforeN = LocalDateTime.now()
val usingMapN = (getA, getB, getC).mapN(add)
usingMapN onComplete {
case Failure(ex) => println(s"Error: ${ex.getMessage()}")
case Success(value) =>
val afterN = LocalDateTime.now()
println(s"beforeN: $beforeN")
println(s"afterN: $afterN")
val diff = getDiff(beforeN, afterN)
println(s"diffN: $diff")
println(s"sum: $value")
}
def getA: Future[Int] = {
println("inside A")
Thread.sleep(3000)
Future.successful(2)
}
def getB: Future[Int] = {
println("inside B")
Thread.sleep(3000)
Future.successful(3)
}
def getC: Future[Int] = {
println("inside C")
Thread.sleep(3000)
Future.successful(4)
}
def add(a: Int, b: Int, c: Int) = a + b + c
def getDiff(before: LocalDateTime, after: LocalDateTime): Long = {
Duration.between(before.toInstant(ZoneOffset.UTC), after.toInstant(ZoneOffset.UTC)).toMillis()
}
}

Answer 1

Because you have sleep outside Future it should be like:
def getA: Future[Int] = Future {
println("inside A")
Thread.sleep(3000)
2
}
So you start async Future with apply - Future.successful on the other hand returns pure value, meaning you execute sleep in same thread.

Answer 2

The time is going before mapN is ever called.
(getA, getB, getC).mapN(add)
This expression is creating a tuple (sequentially) and then calling mapN on it. So it is calling getA then getB then getC and since each of them has a 3 second delay, it takes 9 seconds.